codeforce 6C - Alice, Bob and Chocolate

最新推荐文章于 2021-02-03 19:41:21 发布

这个做不了。

最新推荐文章于 2021-02-03 19:41:21 发布

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分类专栏：大学竞赛文章标签：水题 ACM codeforces 模拟算法

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Alice 和 Bob 进行一场特别的游戏——吃巧克力。他们面前摆放了一排巧克力棒，从两端同时开始食用，直到相遇为止。每块巧克力需要特定的时间才能吃完，且两人进食速度相同。若同时到达同一块巧克力，则 Bob 礼让 Alice。本篇文章将通过模拟的方式解决这个问题，并给出 AC 代码。

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Alice and Bob like games. And now they are ready to start a new game. They have placed n chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.

How many bars each of the players will consume?

Input

The first line contains one integer n (1 ≤ n ≤ 105) — the amount of bars on the table. The second line contains a sequence t1, t2, ..., tn(1 ≤ ti ≤ 1000), where ti is the time (in seconds) needed to consume the i-th bar (in the order from left to right).

Output

Print two numbers a and b, where a is the amount of bars consumed by Alice, and b is the amount of bars consumed by Bob.

一个左边开始，一个右边开始，模拟一下即可

AC代码：

#include<stdio.h>
using namespace std;
typedef long long int ll;
int a[100010], n, sum;
int main() {
	int i, j, k;
	int l=0, r=0, cnt1=0, cnt2=0;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (i = 1, j = n; i <= j; ) {
		if (l <= r) {
			l = l + a[i];
			cnt1++;
			i++;
		}
		else {
			r = r + a[j];
			cnt2++;
			j--;
		}
	}
	printf("%d %d", cnt1, cnt2);
	return 0;
}